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anonymous

  • one year ago

Simplify the rational expression. State any restrictions on the variable.

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  1. anonymous
    • one year ago
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    \[\frac{ n^4-10n^2+24 }{ n^4-9^2+18 }\]

  2. math&ing001
    • one year ago
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    First lets find the restrictions. For that we solve for n : n^4 - 9^2 + 18 = 0

  3. anonymous
    • one year ago
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    Than what? @math&ing001

  4. anonymous
    • one year ago
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  5. anonymous
    • one year ago
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  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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  8. anonymous
    • one year ago
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    @Hero

  9. anonymous
    • one year ago
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    @satellite73

  10. Hero
    • one year ago
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    Let \(n^2 = y\) then try simplifying the expression.

  11. anonymous
    • one year ago
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    Please help me I know its \[\frac{ n^2-4 }{ n^2-3}\] but i dont know the other half for my answer, so it's between A-C for the answer

  12. anonymous
    • one year ago
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    (n+2)(n−2)(n2−6) (n2−3)(n2−6)

  13. Hero
    • one year ago
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    How do you know it's \(\dfrac{n^2 - 4}{n^2 - 3}\)?

  14. anonymous
    • one year ago
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    (n+2)(n−2)(n2−6) (n2−3)(n2−6)

  15. Hero
    • one year ago
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    Multiply it back out to see if what you get matches the original expression.

  16. Hero
    • one year ago
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    Also the original expression you posted likely has a typo in it somewhere.

  17. anonymous
    • one year ago
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    How do i do that?

  18. anonymous
    • one year ago
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    @timo86m

  19. anonymous
    • one year ago
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    @Astrophysics

  20. anonymous
    • one year ago
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    @Michele_Laino can you please help me finish my math problem?

  21. anonymous
    • one year ago
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    @mathmate

  22. anonymous
    • one year ago
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    I just need help with the end of the problem. I think the answer is B though

  23. mathmate
    • one year ago
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    @EllenJaz17 Please confirm that the problem is indeed: \(\Large \frac{ n^4-10n^2+24 }{ n^4-9\color{red}{n}^2+18 } \)

  24. anonymous
    • one year ago
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    yes

  25. anonymous
    • one year ago
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    Yes it is @mathmate

  26. Michele_Laino
    • one year ago
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    hint: we can factorize the denominator, using the same procedure for numerator, so we can write this: \[{n^4} - 9{n^2} + 18 = \left( {{n^2} - 6} \right)\left( {{n^2} - 3} \right)\]

  27. anonymous
    • one year ago
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    (n+2)(n−2)(n2−6) (n2−3)(n2−6)

  28. anonymous
    • one year ago
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    \[\frac{ n^2-4 }{ n^2-3 }\]

  29. Michele_Laino
    • one year ago
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    yes!

  30. anonymous
    • one year ago
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    I know thats the answer to the first half but i need to find out the second half

  31. anonymous
    • one year ago
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  32. anonymous
    • one year ago
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  33. anonymous
    • one year ago
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  34. anonymous
    • one year ago
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  35. Michele_Laino
    • one year ago
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    the denominator is equal to zero when these two conditions hold: \[\begin{gathered} {n^2} - 6 = 0 \hfill \\ \hfill \\ {n^2} - 3 = 0 \hfill \\ \end{gathered} \]

  36. Michele_Laino
    • one year ago
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    please remember, that we can not divide by zero

  37. anonymous
    • one year ago
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    correct

  38. Michele_Laino
    • one year ago
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    please solve those quadratic equations for n, what do you get?

  39. anonymous
    • one year ago
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    6 and 3?

  40. Michele_Laino
    • one year ago
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    are you sure? we have 6^2= 36, and 3^2=9

  41. anonymous
    • one year ago
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    No im not sure im just guessing

  42. Michele_Laino
    • one year ago
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    hint: the solution to this equation: \[{n^2} - k = 0\] are: \[n = \sqrt k ,\quad n = - \sqrt k \]

  43. Michele_Laino
    • one year ago
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    solutions*

  44. Michele_Laino
    • one year ago
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    since: \[{\left( {\sqrt k } \right)^2} = {\left( { - \sqrt k } \right)^2} = {n^2}\]

  45. Michele_Laino
    • one year ago
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    being k a positive number or k=0

  46. anonymous
    • one year ago
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    So is it answer B?

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  47. Michele_Laino
    • one year ago
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    no, I don't think so, sorry!

  48. anonymous
    • one year ago
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  49. Michele_Laino
    • one year ago
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    hint: what are the solution of this equation: \[{n^2} - 3 = 0\] ?

  50. anonymous
    • one year ago
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    So that only leaves A & C correct? because D isn't \[\frac{ n^2-4 }{ n^2-3 }\]

  51. Michele_Laino
    • one year ago
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    solutions*

  52. anonymous
    • one year ago
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    \[n^{2}=3?\]

  53. Michele_Laino
    • one year ago
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    yes! and what is n=...?

  54. anonymous
    • one year ago
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    \[n \neq \pm \sqrt{6}\pm \sqrt{3}?\]

  55. Michele_Laino
    • one year ago
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    from this equation: \[{n^2} - 3 = 0\] I get \[n = \pm \sqrt 3 \] am I right?

  56. anonymous
    • one year ago
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    yes

  57. Michele_Laino
    • one year ago
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    ok! so we have to exclude those 2 values, since when: \[n = \pm \sqrt 3 \] the denominator is zero Now do the same with the equation: \[{n^2} - 6 = 0\]

  58. anonymous
    • one year ago
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    \[n=\pm \sqrt{6}\]

  59. Michele_Laino
    • one year ago
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    perfect, we have to exclude those values too, since when: \[n = \pm \sqrt 6 \] the denominator is zero. So what is the right option?

  60. anonymous
    • one year ago
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    A

  61. Michele_Laino
    • one year ago
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    that's right! :)

  62. anonymous
    • one year ago
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    Thank you so much for teaching me how to get to the answer!

  63. Michele_Laino
    • one year ago
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    :)

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